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I would solve for y = (2-x^2)/x from the second equation and substitute into the first equation to get x^2 - (2-x^2)^2/x^2 - 1 = 0 or x^4 - x^4 + 4x^2 - 4 - x^2 = 3x^2 - 4 = 0, which has roots x = ±2√3/3 and y = ±√(4/3-3/3) = ±√3/3
Друг! Кто тебя научил так коряво решать системы? Умножаешь второе уравнение на 2,вычитаешь первое.Получаешь (x+y)²=3 Остальное - дело техники . Правда, для тех ,у кого она есть.